On Sun, Feb 21, 2021 at 01:43:03PM -0500, Jeff Layton wrote: > On Sun, 2021-02-21 at 16:52 +0000, Al Viro wrote: > > On Sat, Feb 20, 2021 at 01:32:50AM -0500, Luo Longjun wrote: > > > + list_for_each_entry(bfl, &fl->fl_blocked_requests, fl_blocked_member) > > > + __locks_show(f, bfl, level + 1); > > > > Er... What's the maximal depth, again? Kernel stack is very much finite... > > Ooof, good point. I don't think there is a maximal depth on the tree > itself. If you do want to do something like this, then you'd need to > impose a hard limit on the recursion somehow.
I think all you need to do is something like: follow the first entry of fl_blocked_requests, printing as you go, until you get down to lock with empty fl_blocked_requests (a leaf of the tree). When you get to a leaf, print, then follow fl_blocker back up and look for your parent's next sibling on its fl_blocked_requests list. If there are no more siblings, continue up to your grandparent, etc. It's the traverse-a-maze-by-always-turning-left algorithm applied to a tree. I think we do it elsewhere in the VFS. You also need an integer that keeps track of your current indent depth. But you don't need a stack. ? --b.
